Out of the four choices, it would be A
What are you trying to do here?
Solve the graph, or make it appear as something else?
First, we're going to take one sec (x) out so that we get:
sec (x) (2sec (x) -1 -1) = 0
sec (x) (2sec (x) -2) = 0
Then we're going to separate the two to find the zeros of each because anything time 0 is zero.
sec(x) = 0
2sec (x) - 2 = 0
Now, let's simplify the second one as the first one is already.
Add 2 to both sides:
2sec (x) = 2
Divide by 3 on both sides:
sec (x) = 1
I forgot my unit circle, so you'd have to do that by yourself. Hopefully, I helped a bit though!
Based on the z-score, an Endure All battery selected at random has a 10.56% chance of lasting more than 550 hours.
<h3>
What are z-scores?</h3>
How closely a value relates to the mean of a set of values is expressed by a Z-score.
The Z-score is created using standard deviations from the mean.
A data point's total value is equal to the mean value when the Z-score of the data point is equal to 0.
So, in order to determine how far the raw score deviates from the mean, the Z score is used.
The following steps are used to determine the Z score:
z = (x - μ)/σ
We know that: x > 550 and μ = 500, σ = 40
z = 550 - 500 / 40
z = 1.25
The graph of the normal distribution indicates:
P(z > 1.25) = 1 - P(z < 1.25)
P(z > 1.25) = 1 - 0.8944
P(z > 1.25) = 0.1056
Therefore, based on the z-score, an Endure All battery selected at random has a 10.56% chance of lasting more than 550 hours.
Know more about the z-scores here:
brainly.com/question/25638875
#SPJ4
The answer to this would be C because you can multiply everything by 3 and still have a consistent-dependent system.
Answer:
e. The probability of observing a sample mean of 5.11 or less, or of 5.29 or more, is 0.018 if the true mean is 5.2.
Step-by-step explanation:
We have a two-tailed one sample t-test.
The null hypothesis claims that the pH is not significantly different from 5.2.
The alternative hypothesis is that the mean pH is significantly different from 5.2.
The sample mean pH is 5.11, with a sample size of n=50.
The P-value of the test is 0.018.
This P-value corresponds to the probability of observing a sample mean of 5.11 or less, given that the population is defined by the null hypothesis (mean=5.2).
As this test is two-tailed, it also includes the probability of the other tail. That is the probability of observing a sample with mean 5.29 or more (0.09 or more from the population mean).
Then, we can say that, if the true mean is 5.2, there is a probability P=0.018 of observing a sample of size n=50 with a sample mean with a difference bigger than 0.09 from the population mean of the null hypothesis (5.11 or less or 5.29 or more).
The right answer is e.
